Is the closed unit ball of the Hilbert space (or, for that matter, of the Hilbert cube, in some metric) homeomorphic to the unit sphere (viz., its own boundary) ? This is clearly uncharacteristic of finite-dimensional cubes. This question is motivated by general considerations in dimension theory. If there is such a homeomorphism, the small inductive dimension, generalised verbatim to infinite cardinals, cannot exist for such spaces (whose "dimension" is a "strange" cardinal like w).

$\begingroup$Welcome here, interesting question. A remark: You can always edit your own questions and it is better to do so. It is a bit confusing to have both questions here.$\endgroup$
– András BátkaiFeb 13 '13 at 8:18

$\begingroup$Apologies for the duplication in question, more the so after the answer by Martin which showed that the original question was as relevent and, in some sense, equivalent.$\endgroup$
– N UnnikrishnanFeb 23 '13 at 11:00

$\begingroup$The fact that 2 and 3 are homeomorphic is easier than this $\mathbb R^{\mathbb N}$ result. And can be found in more elementary textbooks.$\endgroup$
– Gerald EdgarFeb 13 '13 at 14:33

$\begingroup$On the other hand, if you are interested in this area, then you cannot find a better place than Bessaga & Pelczynski to learn it.$\endgroup$
– Gerald EdgarFeb 13 '13 at 14:35

$\begingroup$Can we expect this phenomenon to happen for every limit ordinal? To be precise, does every R^α contain open sets which are homeomorphic to their boundaries whenever α is a limit ordinal? Or does regularity of α have a role there? Unfortunately I am not able to procure Bessaga & Pelczynski.$\endgroup$
– N UnnikrishnanFeb 23 '13 at 19:37